If alpha and beta are two distinct negative r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given equation: $x^5-5x^3+5x^2-1=0$. By inspection,$x=1$ is a root. Factoring the polynomial: $(x-1)(x^4+x^3-4x^2+x+1)=0$. Further factoring: $(x-

Vedclass · Accepted Answer

$x^4-3x^2+1=0$

Vedclass · Answer

(D) Given equation: $x^5-5x^3+5x^2-1=0$. By inspection,$x=1$ is a root. Factoring the polynomial: $(x-1)(x^4+x^3-4x^2+x+1)=0$. Further factoring: $(x-1)^2(x^3+2x^2-2x-1)=0$. $(x-1)^3(x^2+3x+1)=0$. The roots are $x=1$ (triple root) and $x=\frac{-3 \pm \sqrt{5}}{2}$. The two distinct negative roots are $\alpha = \frac{-3+\sqrt{5}}{2}$ and $\beta = \frac{-3-\sqrt{5}}{2}$. We need an equation with roots $\sqrt{-\alpha}$ and $\sqrt{-\beta}$. Let $y = \sqrt{-\alpha} \Rightarrow y^2 = -\alpha$. Since $\alpha$ and $\beta$ are roots of $x^2+3x+1=0$,we have $\alpha+\beta = -3$ and $\alpha\beta = 1$. Let the new roots be $y_1 = \sqrt{-\alpha}$ and $y_2 = \sqrt{-\beta}$. The equation is $(y^2+\alpha)(y^2+\beta) = 0$. $y^4 + (\alpha+\beta)y^2 + \alpha\beta = 0$. Substituting the values: $y^4 - 3y^2 + 1 = 0$.

If $\alpha$ and $\beta$ are two distinct negative roots of $x^5-5x^3+5x^2-1=0$,then the equation of least degree with integer coefficients having $\sqrt{-\alpha}$ and $\sqrt{-\beta}$ as its roots is

Similar Questions

Let $a, b, c$ and $d$ be any four real numbers. Then $a^{n} + b^{n} = c^{n} + d^{n}$ holds for any natural number $n$ if:

If $\alpha$ is a root of the quadratic equation $x^2 + 6x - 2 = 0$,then another root $\beta$ is

$E_1: a+b+c=0$,if $1$ is a root of $ax^2+bx+c=0$. $E_2: b^2-a^2=2ac$,if $\sin \theta, \cos \theta$ are the roots of $ax^2+bx+c=0$. Which of the following is true?

Evaluate $\sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}}$ for $x \ge 1$.

The sum of the roots of the equation $e^{4t} - 10e^{3t} + 29e^{2t} - 20e^t + 4 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module